Solve (x)^(2)-(10+3x)^(2)=20-x(10+3x) | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$${ x }^{ 2 } - { \left(10+3x \right) }^{ 2 } =20-x(10+3x)$$

Solve for x

$x=-6$
$x=-4$

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring By Grouping

Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(10+3x\right)^{2}$.

$$x^{2}-\left(100+60x+9x^{2}\right)=20-x\left(10+3x\right)$$

To find the opposite of $100+60x+9x^{2}$, find the opposite of each term.

$$x^{2}-100-60x-9x^{2}=20-x\left(10+3x\right)$$

Combine $x^{2}$ and $-9x^{2}$ to get $-8x^{2}$.

$$-8x^{2}-100-60x=20-x\left(10+3x\right)$$

Use the distributive property to multiply $x$ by $10+3x$.

$$-8x^{2}-100-60x=20-\left(10x+3x^{2}\right)$$

To find the opposite of $10x+3x^{2}$, find the opposite of each term.

$$-8x^{2}-100-60x=20-10x-3x^{2}$$

Subtract $20$ from both sides.

$$-8x^{2}-100-60x-20=-10x-3x^{2}$$

Subtract $20$ from $-100$ to get $-120$.

$$-8x^{2}-120-60x=-10x-3x^{2}$$

Add $10x$ to both sides.

$$-8x^{2}-120-60x+10x=-3x^{2}$$

Combine $-60x$ and $10x$ to get $-50x$.

$$-8x^{2}-120-50x=-3x^{2}$$

Add $3x^{2}$ to both sides.

$$-8x^{2}-120-50x+3x^{2}=0$$

Combine $-8x^{2}$ and $3x^{2}$ to get $-5x^{2}$.

$$-5x^{2}-120-50x=0$$

Divide both sides by $5$.

$$-x^{2}-24-10x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}-10x-24=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $-x^{2}+ax+bx-24$. To find $a$ and $b$, set up a system to be solved.

$$a+b=-10$$ $$ab=-\left(-24\right)=24$$

Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. List all such integer pairs that give product $24$.

$$-1,-24$$ $$-2,-12$$ $$-3,-8$$ $$-4,-6$$

Calculate the sum for each pair.

$$-1-24=-25$$ $$-2-12=-14$$ $$-3-8=-11$$ $$-4-6=-10$$

The solution is the pair that gives sum $-10$.

$$a=-4$$ $$b=-6$$

Rewrite $-x^{2}-10x-24$ as $\left(-x^{2}-4x\right)+\left(-6x-24\right)$.

$$\left(-x^{2}-4x\right)+\left(-6x-24\right)$$

Factor out $x$ in the first and $6$ in the second group.

$$x\left(-x-4\right)+6\left(-x-4\right)$$

Factor out common term $-x-4$ by using distributive property.

$$\left(-x-4\right)\left(x+6\right)$$

To find equation solutions, solve $-x-4=0$ and $x+6=0$.

$$x=-4$$ $$x=-6$$

Steps Using the Quadratic Formula

Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(10+3x\right)^{2}$.

$$x^{2}-\left(100+60x+9x^{2}\right)=20-x\left(10+3x\right)$$

To find the opposite of $100+60x+9x^{2}$, find the opposite of each term.

$$x^{2}-100-60x-9x^{2}=20-x\left(10+3x\right)$$

Combine $x^{2}$ and $-9x^{2}$ to get $-8x^{2}$.

$$-8x^{2}-100-60x=20-x\left(10+3x\right)$$

Use the distributive property to multiply $x$ by $10+3x$.

$$-8x^{2}-100-60x=20-\left(10x+3x^{2}\right)$$

To find the opposite of $10x+3x^{2}$, find the opposite of each term.

$$-8x^{2}-100-60x=20-10x-3x^{2}$$

Subtract $20$ from both sides.

$$-8x^{2}-100-60x-20=-10x-3x^{2}$$

Subtract $20$ from $-100$ to get $-120$.

$$-8x^{2}-120-60x=-10x-3x^{2}$$

Add $10x$ to both sides.

$$-8x^{2}-120-60x+10x=-3x^{2}$$

Combine $-60x$ and $10x$ to get $-50x$.

$$-8x^{2}-120-50x=-3x^{2}$$

Add $3x^{2}$ to both sides.

$$-8x^{2}-120-50x+3x^{2}=0$$

Combine $-8x^{2}$ and $3x^{2}$ to get $-5x^{2}$.

$$-5x^{2}-120-50x=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.

$$-5x^{2}-50x-120=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-5$ for $a$, $-50$ for $b$, and $-120$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$x=\frac{-\left(-50\right)±\sqrt{\left(-50\right)^{2}-4\left(-5\right)\left(-120\right)}}{2\left(-5\right)}$$

Square $-50$.

$$x=\frac{-\left(-50\right)±\sqrt{2500-4\left(-5\right)\left(-120\right)}}{2\left(-5\right)}$$

Multiply $-4$ times $-5$.

$$x=\frac{-\left(-50\right)±\sqrt{2500+20\left(-120\right)}}{2\left(-5\right)}$$

Multiply $20$ times $-120$.

$$x=\frac{-\left(-50\right)±\sqrt{2500-2400}}{2\left(-5\right)}$$

Add $2500$ to $-2400$.

$$x=\frac{-\left(-50\right)±\sqrt{100}}{2\left(-5\right)}$$

Take the square root of $100$.

$$x=\frac{-\left(-50\right)±10}{2\left(-5\right)}$$

The opposite of $-50$ is $50$.

$$x=\frac{50±10}{2\left(-5\right)}$$

Multiply $2$ times $-5$.

$$x=\frac{50±10}{-10}$$

Now solve the equation $x=\frac{50±10}{-10}$ when $±$ is plus. Add $50$ to $10$.

$$x=\frac{60}{-10}$$

Divide $60$ by $-10$.

$$x=-6$$

Now solve the equation $x=\frac{50±10}{-10}$ when $±$ is minus. Subtract $10$ from $50$.

$$x=\frac{40}{-10}$$

Divide $40$ by $-10$.

$$x=-4$$

The equation is now solved.

$$x=-6$$ $$x=-4$$

Steps for Completing the Square

Use binomial theorem $\left(a+b\right)^{2}=a^{2}+2ab+b^{2}$ to expand $\left(10+3x\right)^{2}$.

$$x^{2}-\left(100+60x+9x^{2}\right)=20-x\left(10+3x\right)$$

To find the opposite of $100+60x+9x^{2}$, find the opposite of each term.

$$x^{2}-100-60x-9x^{2}=20-x\left(10+3x\right)$$

Combine $x^{2}$ and $-9x^{2}$ to get $-8x^{2}$.

$$-8x^{2}-100-60x=20-x\left(10+3x\right)$$

Use the distributive property to multiply $x$ by $10+3x$.

$$-8x^{2}-100-60x=20-\left(10x+3x^{2}\right)$$

To find the opposite of $10x+3x^{2}$, find the opposite of each term.

$$-8x^{2}-100-60x=20-10x-3x^{2}$$

Add $10x$ to both sides.

$$-8x^{2}-100-60x+10x=20-3x^{2}$$

Combine $-60x$ and $10x$ to get $-50x$.

$$-8x^{2}-100-50x=20-3x^{2}$$

Add $3x^{2}$ to both sides.

$$-8x^{2}-100-50x+3x^{2}=20$$

Combine $-8x^{2}$ and $3x^{2}$ to get $-5x^{2}$.

$$-5x^{2}-100-50x=20$$

Add $100$ to both sides.

$$-5x^{2}-50x=20+100$$

Add $20$ and $100$ to get $120$.

$$-5x^{2}-50x=120$$

Divide both sides by $-5$.

$$\frac{-5x^{2}-50x}{-5}=\frac{120}{-5}$$

Dividing by $-5$ undoes the multiplication by $-5$.

$$x^{2}+\left(-\frac{50}{-5}\right)x=\frac{120}{-5}$$

Divide $-50$ by $-5$.

$$x^{2}+10x=\frac{120}{-5}$$

Divide $120$ by $-5$.

$$x^{2}+10x=-24$$

Divide $10$, the coefficient of the $x$ term, by $2$ to get $5$. Then add the square of $5$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+10x+5^{2}=-24+5^{2}$$

Square $5$.

$$x^{2}+10x+25=-24+25$$

Add $-24$ to $25$.

$$x^{2}+10x+25=1$$

Factor $x^{2}+10x+25$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(x+5\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{1}$$

Simplify.

$$x+5=1$$ $$x+5=-1$$

Subtract $5$ from both sides of the equation.

$$x=-4$$ $$x=-6$$